How to show that $1,\alpha,\alpha^2/2$ is an integral basis of $R=\mathcal{O}\cap \mathbb{Q}[\alpha]$. ($\mathcal{O}$ is ring of algebraic integers)
$\alpha$ is a root of $f(x)=x^3+2x^2+4$ which is irreducible in $\mathbb{Q}(x).$
$\alpha^2$ is a root of $g(x)=x^3-4x^2-16x-16$ which is irreducible in $\mathbb{Q}(x).$
I found the discriminant as $disc(\alpha)=-16.5.7$
Also, I know $\alpha^2/2 \in R$ but $\alpha^2/4 \notin R$ and if $\frac{a+b\alpha}{2}\in R$ with $a,b\in \mathbb{Z}$ then a and b are both even.
But, then how to show that $1,\alpha,\alpha^2/2$ is an integral basis of $R$.
Best Answer
Let $K/\Bbb Q$ be a finite field extension and $R$ be the ring of integers of $K$. In order to show that a given $\Bbb Q$-Basis $(\alpha_1, ...,\alpha_n)$ of $K$ whose elements are already in $R$ is an integral basis, there are a couple of useful tricks, saving us from exhausting calculation of traces: Let $d_K$ be the discriminant of $K/\Bbb Q$ Then:
Lemma 1: $d(\alpha_1,...,\alpha_n)$ = $c^2 d_K$ for some $c\in \Bbb Z$.
Lemma 2: if $d(\alpha_1,...,\alpha_n)=d_K$, then $(\alpha_1,...,\alpha_n)$ is an integral basis.
Lemma 3: $d_K \equiv 0,1 \mod 4$
Lemma 4: If $\sigma_1,...\sigma_n$ are the homomorphisms $K\rightarrow \Bbb C$, then $d(\alpha_1,...,\alpha_n) = \det((\sigma_i \alpha_j)_{i,j})^2$
So let's consider the basis you gave: $(1,\alpha,\alpha^2/2)$ is a $\Bbb Q$-Basis that is already in $R$. Fine. We know that $d(1,\alpha,\alpha^2)$ coincides with the discriminant $\Delta f=2^4\cdot 5\cdot 7$ of $f$. Also, by Lemma 4 the discriminant has some linearity in its arguments, more precisely: $$d(1,\alpha,\frac{\alpha^2}{2})=\frac{1}{4}d(1,\alpha,\alpha^2)=\frac{1}{4}\Delta f=2^2\cdot 5 \cdot 7.$$ By Lemma 1 we have $d_K=2^2 \cdot 5\cdot 7$ or $d_K = 5\cdot 7$. But the last is a contradiction to Lemma 3, so $$d_K=2^2 \cdot 5\cdot 7 = d(1,\alpha,\frac{\alpha^2}{2}).$$ By Lemma 2, we conclude that $(1,\alpha,\alpha^2/2)$ is an integral basis, which completes the proof.